Answer:
-6 < x ≤ 6
Step-by-step explanation:
Neither jajsjxkkskxldodkkdosodjdjdjd
Hi I’m Charles from the University of Alaska Mathematics Department. I have a degree in dentistry, chemistry, zoology, algebra, geometry and finally literature. Anyways I was very interested by your question so I took a look and saw that this indeed is a hard question. However when I look at it it doesnt look to bad but it could pose to be quite difficult. So assuming from the question I could tell that you need to define the square root untill you hit the zero degree of -2. But you could instead solve for negative externalities of the equations and then use the Calvin formula to solve this question. So ultimately the answer is A. I hope this helps!
Answer:
The quadratic x2 − 5x + 6 factors as (x − 2)(x − 3). Hence the equation x2 − 5x + 6 = 0
has solutions x = 2 and x = 3.
Similarly we can factor the cubic x3 − 6x2 + 11x − 6 as (x − 1)(x − 2)(x − 3), which enables us to show that the solutions of x3 − 6x2 + 11x − 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.
Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or more polynomials together to obtain another polynomial. Just as we can divide one whole number by another, producing a quotient and remainder, we can divide one polynomial by another and obtain a quotient and remainder, which are also polynomials.
A quadratic equation of the form ax2 + bx + c has either 0, 1 or 2 solutions, depending on whether the discriminant is negative, zero or positive. The number of solutions of the this equation assisted us in drawing the graph of the quadratic function y = ax2 + bx + c. Similarly, information about the roots of a polynomial equation enables us to give a rough sketch of the corresponding polynomial function.
As well as being intrinsically interesting objects, polynomials have important applications in the real world. One such application to error-correcting codes is discussed in the Appendix to this module.
